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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A die is thrown three times. Let X be ‘the number of twos seen’. Find the expectation of X.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, w e h a v e X = 0, 1, 2, 3 [? d i e i s t h r o w n 3 t i m e s] \\ a n d p = \frac{1}{6}, q = 1 - \frac{1}{6} = \frac{5}{6} \\ ∴ P (X = 0) = P (n o t 2) . P (n o t 2) . P (n o t 2) = \frac{5}{6} . \frac{5}{6} . \frac{5}{6} = \frac{125}{216} \\ P (X = 1) = P (2) . P (2) . P (n o t 2) + P (2) . P (n o t 2) . P (2) + P (n o t 2) . P (2) . P (2) \\ = \frac{1}{6} . \frac{1}{6} . \frac{5}{6} + \frac{1}{6} . \frac{5}{6} . \frac{1}{6} + \frac{5}{6} . \frac{1}{6} . \frac{1}{6} = \frac{5}{216} + \frac{5}{216} + \frac{5}{216} = \frac{15}{216} \\ P (X = 3) = P (2) . P (2) . P (2) = \frac{1}{6} . \frac{1}{6} . \frac{1}{6} = \frac{1}{216} \\ N o w, E (X) = \sum_{i = 1}^{n} p_{i} x_{i} \\ = 0 \times \frac{125}{216} + 1 \times \frac{75}{216} + 2 \times \frac{15}{216} + 3 \times \frac{1}{216} \\ = 0 + \frac{75}{216} + \frac{30}{216} + \frac{3}{216} = \frac{75 + 30 + 3}{216} = \frac{108}{216} = \frac{1}{2} \\ H e n c e, t h e r e q u i r e d i s \frac{1}{2} . \end{array}$

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New question posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

<p><strong> A die is thrown three times. Let X be ‘the number of twos seen’. Find the expectation of X.</strong></p>

7 Views|Posted a year ago

Asked by Alok Kumar Singh

New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A biased die is such that $P (4) = \frac{1}{10}$ and other scores being equally likely. The die is tossed twice. If X is the ‘number of fours seen’, find the variance of the random variable X.

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

For the following probability distribution, determine the standard deviation of the random variable X.

X

2

3

4

P(X)

0.2

0.5

0.3

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol: $\begin{array}{l} \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

The probability distribution of a random variable X is given below:

X

0

1

2

3

P(X)

k

i. Determine the value of k.

ii. Determine P $(x \leq 2)$ and P $(x > 2)$ .

iii. Find $P(x \leq 2) +P (x > 2)$ .

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) W e k n o w t h a t P (0) + P (1) + P (2) + P (3) = 1 \\ \Rightarrow k + \frac{k}{2} + \frac{k}{4} + \frac{k}{8} = 1 \\ \Rightarrow \frac{8 k + 4 k + 2 k + k}{8} = 1 \Rightarrow 15 k = 8 \\ ∴ k = \frac{8}{15} \\ H e r e, n = 8, p = \frac{1}{10}, q = 1 - \frac{1}{10} = \frac{9}{10} \\ (i i) P (X \leq 2) = P (X = 0) + P (X = 1) + P (X = 2) \\ = k + \frac{k}{2} + \frac{k}{4} = \frac{7 k}{4} = \frac{7}{4} \times \frac{8}{15} = \frac{14}{15} \\ a n d P (X > 2) = P (X = 3) = \frac{k}{8} = \frac{1}{8} \times \frac{8}{15} = \frac{1}{15} \\ (i i i) P (X \leq 2) + P (X > 2) = \frac{14}{15} + \frac{1}{15} = \frac{15}{15} = 1 . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Consider the probability distribution of a random variable X:

X

0

1

2

3

4

P(X)

0.1

0.25

0.3

0.2

0.15

Calculate:(i) $V (\frac{X}{2})$ (ii) Variance of X.

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, what is the probability that there will be at least one defective watch?

13 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, n = 7, p = 0.25 = \frac{25}{100} = \frac{1}{4}, q = 1 - \frac{1}{4} = \frac{3}{4} \\ ∴ P (X \geq 2) = 1 - [P (x = 0) + P (x = 1)] \\ = 1 - [{}_{7}{C_{0}} {(\frac{1}{4})}^{0} {(\frac{3}{4})}^{7} + {}_{7}{C_{1}} {(\frac{1}{4})}^{1} {(\frac{3}{4})}^{6}] \\ = 1 - [{(\frac{3}{4})}^{7} + \frac{7}{4} {(\frac{3}{4})}^{6}] = 1 - {(\frac{3}{4})}^{6} [\frac{3}{4} + \frac{7}{4}] \\ = 1 - {(\frac{3}{4})}^{6} (\frac{10}{4}) = 1 - \frac{729}{4096} \times \frac{10}{4} = 1 - \frac{7290}{16384} \\ = \frac{16384 - 7290}{16384} = \frac{9094}{16384} = \frac{4547}{8192} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{4547}{8192} . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Ten coins are tossed. What is the probability of getting at least 8 heads?

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.

12 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, p = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2} \\ \Rightarrow q = 1 - \frac{1}{2} = \frac{1}{2} a n d n = 5 \\ ∴ P (x = r) = {}_{n}{C_{r}} p^{r} q^{n - r} \\ = {}_{5}{C_{3}} {(\frac{1}{2})}^{3} {(\frac{1}{2})}^{5 - 3} = \frac{5!}{3! 2!} . {(\frac{1}{2})}^{3} . {(\frac{1}{2})}^{2} = 10 . \frac{1}{8} . \frac{1}{4} = \frac{5}{16} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{5}{16} . \end{array}$

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X	2	3	4
P(X)	0.2	0.5	0.3

X	0	1	2	3	4
P(X)	0.1	0.25	0.3	0.2	0.15

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